ArrayPartitionI.java

/* (C) 2024 YourCompanyName */
package array;

import java.util.*;

/**
 * Created by gouthamvidyapradhan on 14/08/2019 Given an array of 2n integers, your task is to group
 * these integers into n pairs of integer, say (a1, b1), (a2, b2), ..., (an, bn) which makes sum of
 * min(ai, bi) for all i from 1 to n as large as possible.
 *
 * <p>Example 1: Input: [1,4,3,2]
 *
 * <p>Output: 4 Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4).
 * Note: n is a positive integer, which is in the range of [1, 10000]. All the integers in the array
 * will be in the range of [-10000, 10000].
 *
 * <p>Solution: O(n log n) General idea is to pair the smallest with the next smallest value inorder
 * to get the max sum of minimum.
 */
public class ArrayPartitionI {
  public static void main(String[] args) {
    int[] A = {1, 2, 3, 4};
    System.out.println(new ArrayPartitionI().arrayPairSum(A));
  }

  public int arrayPairSum(int[] nums) {
    Arrays.sort(nums);
    int sum = 0;
    for (int i = 1; i < nums.length; i += 2) {
      sum += Math.min(nums[i - 1], nums[i]);
    }
    return sum;
  }
}